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For the Taylor Series the remainder is of the form $$R_n = \frac{(x-a)^n}{n!} f^{(n)}(\xi) $$ with $a \leq \xi \leq x$

For the series of $e^x$ about $0$ (that is, the Maclaurin series) the remainder is $R_n = \frac{x^n}{n!}e^{\xi}$ with $0 \leq \xi \leq x$. Now, I have to prove that as $n \to \infty$ this $R_n \to 0$. How do I go about it? My problem is that I know nothing about $x$ and whether it is bigger or smaller than $n$.

In short, how do I prove $$\lim_{n\to\infty}\frac{x^n}{n!}e^{\xi}=0$$

Aritra Das
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2 Answers2

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For $x$ in some interval $[a,b]$, $x^n e^\zeta \leq b^n e^b$. Now $e^b$ is just some constant, and can be ignored. You are now looking at $$ \lim_{n \to \infty} \frac{b^n}{n!},$$ which can be handled in very many ways. For a reference on these, see the answers to the question Prove that $\lim \limits_{n \to \infty} \frac{x^n}{n!} = 0$, $x \in \Bbb R$.

Community
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davidlowryduda
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$e^\xi$ is just a constant. Use Stirling's approximation to write

$$ \lim_{n\to \infty} \frac{x^n}{n!}= \lim_{n\to \infty} \frac{x^n}{\sqrt{2 \pi n}\left(\frac{n}{e}\right)^n}= \frac{1}{\sqrt{2\pi}}\lim_{n\to \infty} \left(\frac{ex}{n}\right)^n\frac{1}{\sqrt{n}} $$ For $n>ex$ the term in the brackets is $<1$ and then its clear that the limit is 0.

yohBS
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